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The largest number that the divisor 4 can be multiplied by without exceeding 5 is 1, so the digit 1 is put above the 5 to start constructing the quotient. Next, the 1 is multiplied by the divisor 4, to obtain the largest whole number that is a multiple of the divisor 4 without exceeding the 5 (4 in this case).
Take each digit of the number (371) in reverse order (173), multiplying them successively by the digits 1, 3, 2, 6, 4, 5, repeating with this sequence of multipliers as long as necessary (1, 3, 2, 6, 4, 5, 1, 3, 2, 6, 4, 5, ...), and adding the products (1×1 + 7×3 + 3×2 = 1 + 21 + 6 = 28). The original number is divisible by 7 if and only if ...
Long division is the standard algorithm used for pen-and-paper division of multi-digit numbers expressed in decimal notation. It shifts gradually from the left to the right end of the dividend, subtracting the largest possible multiple of the divisor (at the digit level) at each stage; the multiples then become the digits of the quotient, and the final difference is then the remainder.
Splitting the dividend into smaller Partial Dividends, then dividing this Partial Dividend by only the left-most digit of the divisor will provide the answer one digit at a time. As you solve each digit of the answer you then subtract Product Pairs (UT pairs) and also NT pairs (Number-Tens) from the Partial Dividend to find the next Partial ...
For example, there are six divisors of 4; they are 1, 2, 4, −1, −2, and −4, but only the positive ones (1, 2, and 4) would usually be mentioned. 1 and −1 divide (are divisors of) every integer. Every integer (and its negation) is a divisor of itself. Integers divisible by 2 are called even, and integers not divisible by 2 are called odd.
Dividing 950 by 4 in a single step would require knowing the multiplication table up to 238 × 4. Instead, the division is reduced to small steps. Starting from the left, enough digits are selected to form a number (called the partial dividend) that is at least 4×1 but smaller than 4×10 (4 being the divisor in this problem). Here, the partial ...